Sect 8.1 Systems of Linear Equations A system of linear equations is where all the equations in a system are linear ( variables raised to the first power)

Sect 8.1 Systems of Linear Equations

A system of linear equations is where all the equations in a system are linear ( variables raised to the first power). Equations are typically written in standard form, Ax + By = C. Three possible graphs.

One solutionIndependent & consistent

NO solutionIndependent & inconsistent

Infinite solutionsDependent & consistent

Ways to solve systems.Graphing, Substitution, and Elimination


Substitution Method. 1. Choose an equation and get x or y by itself.

2. Substitute step 1 equation into the second equation.

3. Solve for the remaining variable.

4 Substitute this answer into the step 1 equation.

Ways to solve systems.Graphing, Substitution, and Elimination

3

1123

yx

yxSolve for x & y.

1123 yx3

3

xy

yx

3xy

)4,1(

4

31

y

y

Is the intersection point and solution.

Step 1 Step 2

Step 3

Step 4

1

55

1165

11623

11323

x

x

x

xx

xx


Elimination Method. 1. Choose variable to cancel out.

Look for opposite signs.

2. Add the equations together to cancel.

3. Solve for the remaining variable.

4 Substitute this answer into either equation in the step 1 equations.

1232

143

yx

yx

Solve for x & y.

(3, 2) is the intersection point and solution.

The y-terms are opposite signs.Multiply the first equation by 3 and the second equation by 4.

4812841232

31293143

yxyx

yxyx

51017 yx

2

63

1236

12332

y

y

y

y

Step 1

Step 2+

Step 3Step 4

317

51

17

17

x

x


646

423

yx

yxSolve for x & y.

24

428

yx

yxSolve for x & y.Elimination

2( )

6x – 4y = 8+

0 = 14

False statement…No Solution

( )/2

Elimination

You can also divide by 2.

4x – y = -2+

0 = 0

True statement…Infinite Solutions!How to write the answers. The solutions is the graph of the line. Convert one of the equations into y = mx + b form.

-4x + y = 2+4x +4x y = 4x + 2

Solution is ( x, 4x +2 )


2

22

3693

zyx

zyx

zyx

Solve for x, y, & z.

Elimination Method with 3 by 3 systems.Step 1. Choose a TERMINATOR equation! Look for coefficients of 1!!

Equation #3.

T

Step 2. Pair this equation together with the other two equations. Also decide which variable to eliminate, it must be the same variable for both pairings!2

2 2

x y z

x y z

2

3 9 6 3

x y z

x y z

Cancel the z terms!+

3x + 2y = 4

-6( )

-6x – 6y – 6z = -12

-3x + 3y = -9 Divide by 3.

-x + y = -3

Step 3. Bring the two new equations together as a 2 by 2 system and solve.

3( )3x + 2y = 4

-3x +3y = -9 +

5y = -5

y = -1 Step 4. Back substitute.

-x – 1 = -3

x = 2

2 zyx

T

212 z1z

1,1,2,, zyx

3x + 2y = 4

-x + y = -3

-x + y = -3

2

0624

032

zyx

zyx

zyxSolve for x, y, & z. Step 1. Choose a

TERMINATOR

T

032

2

zyx

zyx

0624

2

zyx

zyx

Step 2. Pair T with the other two equations. Cancel y’s.

+

3x – 2z = 26x – 4z = 4

2( )

2x – 2y + 2z = 4+

Divide by -2.-3x + 2z = -2Step 3.

0 = 0

+

Infinite solutions.Let z be independent!Solve for x.

3x = 2 + 2z

3

22 zx

2 zyx

23

22

zy

z

zzyx _____,_____,,,

3

22 z

63322 zyzzy 543

3

54 z

Step 4.

42 zyx

2( )943 zyx

18826

42

zyx

zyx

+

1477 zx

2 zx

2zx

zzyx _____,_____,,,

2z

422 zyz622 zy

3 zy 3 z

3x – 2z = 2

-3x + 2z = -2

42 zyx

14

92

22

zyx

zyx

zyx

Solve for x, y, & z.Step 1. Choose a TERMINATOR

T

22

14

zyx

zyx

92

14

zyx

zyx

Step 2. Pair T with the other two equations. Cancel z’s.

+ +

133 yx 822 yx Divide by 2.4 yx

133 yx

4 yx -3( )

1233 yx+

110 False statement…No Solution.


Find the equation of the parabola that passes through the points (2, 4), (-1, 1), and (-2, 5).

cbxaxy 2

We need to find a, b, and c in the equation. Three unknown variables means we need to create three equations. Each point will generate an equation.

424224:4,2 2 cbacba

1111:1,1 2 cbacba

524225:5,2 2 cbacba

424 cba

1 cba

524 cba

T

Cancel the c’s.

1 cba424 cba

1 cba524 cba

+ +

-1( )

333 ba1ba

43 ba

1ba+

54 a25.1

4

5a25.0

125.1

b

b

1 cba

5.0

125.025.1

c

c

5.025.025.1 2 xxy

Sect 8.2 Gauss-Jordan Method to solve Systems of Linear Equations

2

22

3693

zyx

zyx

zyx

Convert a system into an augmented matrix. An augmented matrix is made up rows and columns using only the coefficients on the variables and the constants.

___ ___ ___ ___

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Matrix row transformations.1. Interchange any two rows.2. Multiply or divide the elements of any row by a nonzero real number.3. Replace any row in the matrix by the sum of the elements of that row and a multiple of the elements of another row.

Reduced row echelon form

Steps for the Gauss-Jordan Method.1. Obtain 1 as the first element in the 1st column.2. Use the 1st row to transform the remaining entries in the 1st column to 0.3. Repeat step 1 and 2 by obtaining the 1 as the 2nd element in the 2nd column and use the 2nd row to transform the remaining entries in the 2nd column to 0.4. Repeat step 1 and 2 by obtaining the 1 as the 3rd element in the 3rd column and use the 3rd row to transform the remaining entries in the 3rd column to 0.5. Repeat until the Coefficient matrix becomes the Identity Matrix.

3 9 6 32 1 -1 2

1 1 1 2

1 0 0 2 0 1 0 -1

0 0 1 1

Answers for


1925

143

yx

yx

Solve for x and y.

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1154

1132

yx

yx

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3 - 4 1

5 2 19

1 - 10 -17

5 2 19

2R1 – R2 R1

6-5

-5R1 + R2 R2

-5

1 - 10 -17

0 52 104

R2 / 52 R2

1 -10 -17

0 1 2

10R2 + R1 R1

1 0 3

0 1 2

( x, y ) = ( 3, 2 )

3 2 11

5 4 11

y x

y x

Y X

= Y

= X

1 - 8 -33

5 4 11

___ ___ ___

___ ___ ___

-3 2 11

5 4 11

-2R1 – R2 R1

6-5

-5R1 + R2 R2

-5

1 - 8 -33

0 44 176

R2 / 44 R2

1 - 8 -33

0 1 4

8R2 + R1 R1

+0 1 0 - 1

0 1 4

( x, y ) = ( 4, -1 )

R1

R2

+20

-8-2 2-19

+85+50

+10 +0

R1

R2

+40 +165

-4-4 -22-11

+8 +32


2

22

3693

zyx

zyx

zyxSolve for x, y, & z.

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1 3 2 1

2 1 -1 2

1 1 1 2

- 2R1 + R2 R2

- 1R1 + R3 R3

-2 -6 -4 -2

1 3 2 1

0 -5 - 5 0

0 -2 - 1 1

1 3 2 1

0 1 1 0

0 -2 -1 1

R2 / -5 R2 - 3R2 + R1 R1

2R2 + R3 R3

+0 -3 - 3 +0

+0 +2 +2 +0

1 0 -1 1

0 1 1 0

0 0 1 1

-1R3 + R2 R2

R3 + R1 R1

+0 + 0 +1 + 1 1 0 0 2

0 1 0 - 1

0 0 1 1

+0 +0 -1 - 1

( x, y, z ) = ( 2, -1, 1 )

( )/3R1

R2

R3

Always reduce rows or equations when possible.

Column 1, 1 first. Column 1, 0’s 2nd .

-1 -3 -2 -1

Column 2, 1 first.

Column 2, 0’s 2nd . Column 3, 1 is done, 0’s 2nd .

This process must be done by hand! This process is also programmed into your calculator. Find the Matrix button, for most above x -1 button. There are 3 categories, Names, Math, and Edit. We want Edit 1st. Pick a Matrix and hit Enter, select the dimensions of your Matrix, and enter the data values by rows. DOUBLE CHECK THE ENTRIES, ONE MISTAKE AND THE ANSWER IS WRONG! 2nd Quit for the home screen and go back to the Matrix window for Math. Scroll up to select rref( and hit Enter. Back to Matrix window and stay in the Names window. Select your matrix and hit Enter. Close the parenthesis and Enter.

2

22

3693

zyx

zyx

zyx


Solve for x, y, & z.___ ___ ___ ___

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1 1 1 4

3 4 1 13

2 1 4 7

R1 R3 - 3R1 + R2 R2

1 1 1 4

0 1 - 2 1

0 -1 2 - 1

R2 + R3 R3

-1R2 + R1 R1

+0 -1 +2 -1

1 0 3 3

0 1 -2 1

0 0 0 0

( x, y, z ) = ( 3 – 3z, 1 + 2z, z )

2 4 7

3 4 13

4

x y z

x y z

x y z

- 2R1 + R3 R3

-3 -3 -3 -12

-2 -2 -2 -8

1 1 1 4

0 1 - 2 1

0 -1 2 - 1

+0 + 1 -2 +1

3 3

2 1

x z

y z

3 3

1 2

x z

y z

Infinite Solutions. z is the independent variable.

x y z


Solve for x, y, & z.___ ___ ___ ___

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5 3 9 19

3 4 1 13

3 2 5 12

2R2 – R1 R1

R3 / 13 R3

R2 /-11 R2

1 5 -7 7

0 1 -2 8/11

0 -1 2 -9/13

- 3R1 + R3 R3

6- 8- 2- 26- 1 5 -7 7

0 -11 22 -8

0 -13 26 - 9

5 3 9 19

3 4 13

3 2 5 12

x y z

x y z

x y z

1 5 -7 7

3 4 1 13

3 2 5 12

- 3R1 + R2 R2

-3 - 15 +21 -21

-3 -15 +21 -21

R2 + R3 R3

___ ___ ___ ___

___ ___ ___ ___

___ ___ ___ ___

1 5 -7 7

0 1 -2 8/11

0 0 0 5/143

NO SOLUTION

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Sect 8.1 Systems of Linear Equations A system of linear equations is where all the equations in a system are linear ( variables raised to the first power)